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Let $\Phi$ be a (crystallographic) root system with Weyl group $\mathcal{W}$, and $\Phi^+$ a choice of positive roots, and $$ q := \prod_{\alpha\in\Phi^+} (\exp(\alpha/2) - \exp(-\alpha/2)) = \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w\rho) $$ be the denominator in the Weyl character formula ("WCF"); here, of course, $\exp$ is a formal exponential, $\rho := \frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha = \sum_i\varpi_i$ is the Weyl vector, $\varpi_i$ are the fundamental weights, and $\mathrm{sgn}$ is the abelian character of $\mathcal{W}$ with value $-1$ on the reflections.

Since $q$ is $\mathcal{W}$-anti-invariant (we have $w(q) = \mathrm{sgn}(w)\,q$), it follows that $q^2$ is $\mathcal{W}$-invariant. So $q^2$ can be expressed as a polynomial in the fundamental characters (the fundamental characters being the $x_i := q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ by the WCF); or equivalently, as a polynomial in the averages of the fundamental weights (meaning the $\frac{1}{\#\mathcal{W}} \sum_{w\in\mathcal{W}} \exp(w\varpi_i)$): see the references to Bourbaki and Lorenz in this related question. This invariant quantity $q^2$ can be considered as a kind of discriminant (see PS3 below).

My question about this square denominator $q^2$ is: how can we compute it in practice (as a polynomial of the kind I just described)? Is there a convenient expression? Or perhaps, can I get LiE or Sage to compute it?

I am also interested in any sort of remarks about it: for example, does it have a standard name (beyond "the square of the Weyl denominator")? Is it the (virtual) character of some naturally defined element in the Grothendieck group of representations?

PS 1: I should have made it clear that I am looking for a computational approach that does not involve writing down all the $\#\mathcal{W}$ terms in $q$. (LiE is capable of doing computations on the ring of [multiplicative] $\mathcal{W}$-invariants without fully expanding them; so the question is whether we can do this for a $\mathcal{W}$-anti-invariant like $q$.)

PS 2: If we let $R = \mathbb{C}[\exp(\Lambda)]^{\mathcal{W}}$, where $\Lambda := \langle\Phi^\vee\rangle^*$ is the weight lattice, be the ring of multiplicative $\mathcal{W}$-invariants (which is a polynomial ring), then the ring of multiplicative $\mathcal{W}_0$-invariants, where $\mathcal{W}_0 := \ker(\mathrm{sgn})$ is the group of rotations in the Weyl group, is the free quadratic algebra $R \oplus R q$. So asking to describe $q^2$ in $R$ is the missing bit in the presentation of this algebra. Hopefully this helps motivate the question.

PS 3: In the case where $\Phi = A_n$, then $q^2$ expressed as a polynomial of $x_1,\ldots,x_n$ is exactly the discriminant of the polynomial (in the $z$ variable) $z^{n+1} - x_1 z^n + x_2 z^{n-1} + \cdots + (-1)^n x_n z + (-1)^{n+1}$ (indeed, consider a diagonal element of $SL_{n+1}$: then $x_1,\ldots,x_n$ give the elementary symmetric functions of the $n+1$ eigenvalues, whose product is $1$, and $q^2$ is their discriminant since $q$ is the Vandermonde determinant). What I'm asking for is a generalization of this to the other root systems.

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  • $\begingroup$ I don't understand "$q^2$ can be expressed as a polynomial of the fundamental characters ($q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$)". Are you saying that $q^2$ equals your expression with denominator $q$, so that $q^3$ equals $\sum_{w \in \mathcal W} \operatorname{sgn}(w)\exp(w(\rho + \varpi_i))$? $\endgroup$ – LSpice Jun 1 '17 at 22:43
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    $\begingroup$ @LSpice I'm saying that if we let $x_i:=q^{-1} \sum_w\mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ (the character with highest weight $\varpi_i$: this is just the Weyl character formula), then $q^2$ is some polynomial in $x_1,\ldots,x_n$ where $n$ is the rank of $\Phi$. This does have $q$ in the denominator, but it just means it cancels somehow (that's the whole point of the WCF). $\endgroup$ – Gro-Tsen Jun 2 '17 at 7:32
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    $\begingroup$ For example, for $A_2$, we have $q^2 = x_1^2 x_2^2 - 4 x_1^3 - 4 x_2^3 + 18 x_1 x_2 - 27$ (the standard cubic discriminant when the product of the three roots is $1$) where $x_i = q^{-1} \sum_w \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ and $\rho=\varpi_1+\varpi_2$. $\endgroup$ – Gro-Tsen Jun 2 '17 at 7:35
  • $\begingroup$ Oh, I see; the parenthetical is defining fundamental characters, not exhibiting $q^2$ as such a polynomial. $\endgroup$ – LSpice Jun 2 '17 at 14:02
  • $\begingroup$ @LSpice I agree that my wording was very unclear. I'll try to improve it somewhat. $\endgroup$ – Gro-Tsen Jun 2 '17 at 15:30
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WCR = WeylCharacterRing(['A',3], style='coroots')
WR = WeightRing(WCR)
PR = WR.positive_roots()
q = prod([WR(1/2*x) - WR((-1/2*x)) for x in PR])
qs = q*q
qs.character()

105*A3(0,0,0) - 6*A3(0,2,0) - 15*A3(0,0,4) + 27*A3(0,1,2) - 45*A3(1,0,1) - 12*A3(1,2,1) + 6*A3(1,1,3) + 9*A3(0,4,0) - 3*A3(0,3,2) + 27*A3(2,1,0) - 9*A3(2,0,2) - 15*A3(4,0,0) - 3*A3(2,3,0) + A3(2,2,2) + 6*A3(3,1,1) - 3*A3(3,0,3)

https://cocalc.com/projects/25e624ba-9091-484f-af2e-71deb7120f58/files/Weyl%20denominator%20square.sagews

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  • $\begingroup$ Nice, I didn't know about WeylCharacterRing. However, since I want to compute $q^2$ in the $E_8$ case, this brute force approach can't work (there are $696\,729\,600$ terms in $q$). I'll clarify the question. $\endgroup$ – Gro-Tsen Jun 1 '17 at 21:18
  • $\begingroup$ @Gro-Tsen I see. My code is works perhaps for classical up to rank 4 or 5 and $G_2$. See the updated CoCalc worksheet. By the way, it seems that the resulting polynomial is invariant with respect to some Dynkin diagram automorphisms. $\endgroup$ – Vít Tuček Jun 1 '17 at 22:02
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    $\begingroup$ @VítTuček, of course these symmetries are no accident; the square is $(-1)^{\lvert\Phi\rvert/2}\prod_{\alpha \in \Phi} (\exp(\alpha) - 1)$, which is manifestly symmetric under all automorphisms of $\Phi$. $\endgroup$ – LSpice Jun 1 '17 at 22:35
  • $\begingroup$ @LSpice Of course. I made an off-by-one error when I was looking for the triality symmetry at D3. :) $\endgroup$ – Vít Tuček Jun 1 '17 at 22:39

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